201. A meiotic cell contains fully condensed bivalents and terminalised chiasmata. Its spindle is assembling, while the nucleolus and nuclear envelope are disappearing. This combination identifies:
ⓐ. late diakinesis before metaphase I
ⓑ. pachytene during active crossing over
ⓒ. diplotene before chromosome condensation
ⓓ. early zygotene before synapsis
Correct Answer: late diakinesis before metaphase I
Explanation: Diakinesis completes prophase I. By its later part, the chromosomes are fully condensed, chiasmata have undergone terminalisation and the meiotic spindle is being assembled. The nucleolus disappears, and the nuclear envelope loses its visible integrity, allowing the bivalents to interact with the spindle during the approaching metaphase I. Zygotene would show developing synapsis, and pachytene would contain fully paired tetrads with recombination nodules. Diplotene would show partial homologue separation at visible chiasmata but less advanced terminalisation and nuclear reorganisation. The complete feature set places the cell at the transition from prophase-I chromosome preparation to equatorial bivalent alignment. Stage recognition is strongest when homologue relationship, associated structures and nuclear-envelope status are interpreted together.
202. Match each stage with its characteristic event. Each Column II entry is used once.
| Column I | Column II |
|---|
| P. Pachytene | 1. Bivalents align at the equatorial plate |
| Q. Diplotene | 2. Crossing over occurs at recombination nodules |
| R. Diakinesis | 3. Synaptonemal complex dissolves and chiasmata become visible |
| S. Metaphase I | 4. Chiasmata terminalise and chromosomes become fully condensed |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-2, Q-3, R-4, S-1
Correct Answer: P-2, Q-3, R-4, S-1
Explanation: Pachytene contains fully organised tetrads, recombination nodules and crossing over between non-sister chromatids. Diplotene follows after recombination has been completed; the synaptonemal complex dissolves, homologues begin separating and chiasmata become evident. During diakinesis, chromosomes become fully condensed and chiasmata move towards terminal positions. Metaphase I then places the bivalents at the equatorial plate. The mapping follows a continuous progression from genetic exchange to visible crossover connections, final prophase-I preparation and spindle-dependent alignment. Each stage is distinguished by a different chromosome relationship rather than merely by increasing chromosome condensation. The correct correspondence preserves both the sequence of stages and the identity of the structures involved. Column labels are reference numbers only; their visual position does not imply a biological pairing.
203. A treated meiocyte develops terminalised chiasmata, fully condensed bivalents and an assembled spindle. However, its nucleolus and nuclear envelope remain visibly intact. The treatment most directly disrupted:
ⓐ. crossing over between non-sister chromatids
ⓑ. loss of nuclear structures in late diakinesis
ⓒ. synaptonemal-complex formation in zygotene
ⓓ. homologue separation during anaphase I
Correct Answer: loss of nuclear structures in late diakinesis
Explanation: Terminalised chiasmata and fully condensed bivalents show that the cell has progressed through diplotene into diakinesis. Spindle assembly also indicates preparation for metaphase I. The abnormal observations are the continued visibility of the nucleolus and nuclear envelope, structures that normally disappear by late diakinesis. The treatment affects this specific nuclear reorganisation rather than earlier synapsis or crossing over. Anaphase-I segregation has not yet begun, as the cell remains in the preparatory interval before metaphase alignment. Persistence of the nuclear envelope could interfere with normal interaction between bivalents and the spindle even though chromosome condensation and chiasma terminalisation have occurred. The case demonstrates that several late-diakinesis events can be experimentally separated. The absence of a later product is interpreted through the first failed prerequisite, not as evidence that every earlier event also failed.
204. Consider the following statements about metaphase I.
I. Bivalents align at the equatorial plate.
II. The two homologous chromosomes of a bivalent are connected towards opposite spindle poles.
III. Sister chromatids separate after their centromeres divide.
IV. Centromeres of sister chromatids remain unsplit.
ⓐ. I and III only
ⓑ. II, III and IV only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: During metaphase I, paired homologous chromosomes align as bivalents at the equatorial plate. Spindle microtubules from opposite poles interact with the kinetochores of the two homologues, preparing them to move into different daughter regions. The sister chromatids of each chromosome remain joined at their centromere, so statement IV is valid. Statement III describes anaphase II, not metaphase I or anaphase I. In the first meiotic division, homologous chromosomes separate while sister-centromere association is retained. The valid statements define the reductional geometry of metaphase I: the unit aligned at the equator is a homologous pair, and the forthcoming segregation separates homologues rather than sisters. After meiosis I, DNA content per cell is \(2C\), but the summed DNA across both products remains \(4C\) when no DNA is lost.
205. Four dividing cells show the following arrangements.
| Cell | Equatorial unit | Opposite-pole relation | Centromere state |
|---|
| P | Bivalent | One homologue towards each pole | Unsplit |
| Q | Individual replicated chromosome | One sister towards each pole | Unsplit |
| R | Individual chromosome in a haploid cell | One sister towards each pole | Unsplit |
| S | Separated daughter chromosomes | Moving towards opposite poles | Split |
Which cell is specifically in metaphase I?
ⓐ. Cell Q
ⓑ. Cell P
ⓒ. Cell R
ⓓ. Cell S
Correct Answer: Cell P
Explanation: Cell P aligns a bivalent rather than an individual replicated chromosome. Its two homologous members are related to opposite spindle poles, while their centromeres remain unsplit. This is the defining metaphase-I arrangement and prepares homologues for separation in anaphase I. Cell Q represents the general geometry of mitotic metaphase, where sister kinetochores of an individual replicated chromosome face opposite poles. Cell R fits metaphase II in a haploid cell, which also aligns chromosomes individually. Cell S has already entered anaphase after centromere division. The table distinguishes three superficially similar equatorial or spindle arrangements by identifying the aligned unit, the chromosome relationship oriented towards opposite poles and the state of the centromeres. The first biologically meaningful difference among the rows locates the process or stage represented by the data. Row interpretation requires the same counting convention and observation unit to be applied consistently across the table.
206. Consider the following events during anaphase I.
I. Homologous chromosomes move towards opposite poles.
II. Centromeres of sister chromatids remain undivided.
III. Sister chromatids of each chromosome move together.
IV. Each separating chromatid immediately becomes an independent daughter chromosome.
ⓐ. I, II and IV only
ⓑ. II, III and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Anaphase I separates the two homologous chromosomes of each bivalent. One replicated homologue moves towards one pole, while the corresponding homologue moves towards the opposite pole. The centromere of each chromosome remains intact, so its two sister chromatids continue travelling together as one replicated chromosome. Statement IV describes the change occurring in mitotic anaphase and anaphase II, where centromeres split and former sister chromatids become independent daughter chromosomes. The first meiotic division is reductional precisely since homologous chromosome sets, rather than sister chromatids, are separated. Each future daughter nucleus receives one member of every homologous pair while retaining chromosomes composed of two sister chromatids. Bivalent orientation at metaphase I prepares homologue separation; sister-chromatid separation is reserved for the second meiotic division.
207. Use the chromosome movement described here. Replicated homologue P, consisting of sister chromatids P1 and P2, moves towards one pole. Its replicated homologue Q, consisting of Q1 and Q2, moves towards the opposite pole. P1 remains joined to P2, and Q1 remains joined to Q2. The cell is in:
ⓐ. mitotic anaphase
ⓑ. meiotic metaphase II
ⓒ. meiotic anaphase I
ⓓ. meiotic anaphase II
Correct Answer: meiotic anaphase I
Explanation: The separating units are replicated homologous chromosomes. P moves as one chromosome with its two sister chromatids still joined, while Q moves in the opposite direction with its sister pair also intact. This is the defining chromosome behaviour of anaphase I. During mitotic anaphase and anaphase II, the centromeres divide and sister chromatids separate from each other. Metaphase II would show individual replicated chromosomes aligned at an equator rather than moving poleward. The described movement also explains the reductional nature of meiosis I. Each pole receives one homologue from every pair, reducing the chromosome-set number, but the chromosomes arriving there still contain the sister chromatids produced during premeiotic DNA replication.
208. A meiocyte has \(2n=18\). During normal anaphase I, homologues have reached opposite sides, but cytokinesis has not begun and all centromeres remain intact. Each pole contains:
ⓐ. \(18\) chromosomes, \(36\) chromatids and \(4C\) DNA
ⓑ. \(18\) chromosomes, \(18\) chromatids and \(2C\) DNA
ⓒ. \(9\) chromosomes, \(9\) chromatids and \(1C\) DNA
ⓓ. \(9\) chromosomes, \(18\) chromatids and \(2C\) DNA
Correct Answer: \(9\) chromosomes, \(18\) chromatids and \(2C\) DNA
Explanation: A diploid number of \(18\) represents \(9\) homologous pairs. During anaphase I, one replicated chromosome from each pair moves towards each pole. Each pole receives:
\[
\frac{18}{2}=9\ \text{chromosomes}
\]
The centromeres remain intact, so every chromosome still consists of two sister chromatids:
\[
9\times2=18\ \text{chromatids}
\]
The original meiocyte contained \(4C\) DNA after premeiotic replication. Equal homologue segregation places half of that amount, or \(2C\), at each pole. The chromosome set at either pole is haploid, but its chromosomes remain replicated. The condition per pole is \(n,2C\), not \(n,1C\), which is reached only after sister-chromatid separation in meiosis II. Meiosis I is reductional since homologous chromosome sets move to opposite poles while sister centromeres remain intact. Each pole receives one member of every homologous pair, so ploidy falls to \(n\) even though each chromosome still contains two sister chromatids. Bivalent orientation at metaphase I prepares homologue separation; sister-chromatid separation is reserved for the second meiotic division.
209. Control meiocytes align bivalents normally and then move one homologue of each pair to each pole. In treated cells, bivalents align normally, but one homologue of several pairs fails to move after anaphase I begins. The treatment most directly interfered with:
ⓐ. spindle-driven homologue movement in anaphase I
ⓑ. synapsis of homologues during zygotene
ⓒ. premeiotic replication of sister chromatids
ⓓ. nuclear-envelope reformation during telophase II
Correct Answer: spindle-driven homologue movement in anaphase I
Explanation: Normal bivalent alignment shows that homologous pairing, prophase-I organisation and metaphase-I positioning have already occurred. The defect appears only after anaphase I begins, when individual homologues should move towards opposite poles. Failure of selected homologues to migrate points to disruption of the spindle-related movement required for their segregation. A replication defect would have affected chromosome structure before prophase I, while failed synapsis would have prevented normal bivalent formation and alignment. Telophase-II nuclear-envelope formation occurs much later and cannot explain the immediate movement failure. The experiment distinguishes successful equatorial preparation from successful poleward segregation and identifies spindle-dependent homologue movement as a separate requirement of the reductional division. The conclusion remains limited to the supplied treatment and observations; broader causal claims would require additional evidence. The treatment is located by finding the earliest event that differs from the control and then tracing its expected downstream consequence.
210. Assertion: Each chromosome reaching a pole in normal anaphase I still consists of two sister chromatids.
Reason: Centromeres do not divide during anaphase I, so the two sister chromatids of each chromosome move together towards the same pole.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain the Assertion
ⓒ. Assertion is true and Reason is false, so Reason cannot correctly explain the Assertion
ⓓ. Assertion is false and Reason is true, so Reason cannot correctly explain the Assertion
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains the Assertion
Explanation: The assertion is true because anaphase I separates homologous chromosomes rather than sister chromatids. Each chromosome moving towards a pole therefore remains replicated and consists of two sister chromatids. The reason is also true and provides the direct explanation: centromeres do not divide during anaphase I, so the sister chromatids remain physically associated and move together as one chromosome. This behaviour distinguishes anaphase I from mitotic anaphase and anaphase II, in which centromeres divide and sister chromatids separate as daughter chromosomes. Separation of homologous chromosomes reduces the chromosome-set number at each pole from diploid to haploid, while the retention of joined sister chromatids preserves \(2C\) DNA in each resulting cell. The absence of centromere division is therefore the immediate structural basis for each chromosome reaching the pole with both sister chromatids still attached.
211. The observations below were recorded in four dividing cells.
| Cell | Separating or arranged units | Centromere state | Resulting relation |
|---|
| P | Homologous chromosomes separating | Unsplit | One homologue moves towards each pole |
| Q | Sister chromatids separating | Split | One daughter chromosome moves towards each pole |
| R | Bivalents aligned at the equator | Unsplit | Homologue separation has not begun |
| S | Chromosomes decondensing at poles | Separated | Nuclear envelopes are re-forming |
Which pair best contrasts the reductional separation of meiosis I with an equational sister-separation event?
ⓐ. Cells P and R
ⓑ. Cells P and Q
ⓒ. Cells Q and S
ⓓ. Cells R and S
Correct Answer: Cells P and Q
Explanation: Cell P shows the defining event of anaphase I: homologous chromosomes move towards opposite poles while sister centromeres remain intact. Separation of homologous sets reduces the chromosome-set number in each product, so this is the reductional event. Cell Q shows centromere division followed by separation of sister chromatids into daughter chromosomes. That behaviour is equational with respect to ploidy and occurs in mitotic anaphase or anaphase II, depending on the cell's history. Cell R is still at metaphase I, before homologue segregation, while Cell S is undergoing telophase reorganisation after chromosome movement. Comparing P with Q isolates the decisive distinction between the two separation programmes: meiosis I resolves homologous pairs without splitting centromeres, whereas an equational anaphase resolves sister chromatids after centromere division. The retained or divided centromere is the decisive structural clue.
212. Centromeres split prematurely while homologous chromosomes are beginning to segregate in anaphase I. The most immediate consequence would be:
ⓐ. formation of normal bivalents during the same division
ⓑ. restoration of the diploid condition at each pole
ⓒ. completion of meiosis II before telophase I
ⓓ. premature sister-chromatid separation during meiosis I
Correct Answer: premature sister-chromatid separation during meiosis I
Explanation: Normal anaphase I separates homologous chromosomes while preserving the centromeric connection between sister chromatids. Premature centromere division would release the sisters during a stage in which they should move together as one replicated chromosome. The usual division pattern would be disrupted: homologue segregation and sister separation could occur at the same time instead of being assigned to meiosis I and meiosis II respectively. This would not restore diploidy at either pole or create new bivalents. It also would not represent orderly completion of meiosis II, since the second division involves its own spindle organisation and chromosome alignment. The immediate defect is loss of the sister association that normally preserves replicated chromosomes after the reductional division.
213. In a meiocyte where nuclear structures reappear after meiosis I, arrange the following events in their broad order.
P. Homologous chromosomes reach opposite poles.
Q. Chromosomes begin dispersing from their condensed state.
R. Nuclear membranes and nucleoli reappear.
S. Cytokinesis produces two cells forming a dyad.
ⓐ. Q → P → S → R
ⓑ. P → R → S → Q
ⓒ. R → P → Q → S
ⓓ. P → Q → R → S
Correct Answer: P → Q → R → S
Explanation: Homologous chromosome segregation is completed as the replicated homologues reach opposite poles during late anaphase I. Telophase I then begins, and the chromosomes may disperse from their highly condensed condition without necessarily becoming as extended as ordinary interphase chromatin. In the stated meiocyte, nuclear membranes and nucleoli reappear around the chromosome groups. Cytokinesis can subsequently divide the cytoplasm and form two haploid cells, collectively described as a dyad. The broad order follows the transition from chromosome segregation to partial nuclear reorganisation and cellular separation. Some events may overlap, and nuclear reappearance is not equally prominent in every organism, but cytokinesis cannot normally produce the dyad before the homologous chromosome sets have been established. Centromere state distinguishes anaphase I from anaphase II: intact sister centromeres accompany homologue movement in the first division.
214. Consider the following statements about telophase I.
I. Nuclear membranes and nucleoli may reappear.
II. Chromosomes may disperse without becoming fully extended.
III. Cytokinesis may produce a dyad of haploid cells.
IV. Sister chromatids separate after centromere division.
ⓐ. II, III and IV only
ⓑ. I and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Telophase I follows separation of homologous chromosomes. Nuclear membranes and nucleoli may re-form around the chromosome groups, although the extent of nuclear reconstitution varies. The chromosomes may become less condensed, but they often do not reach the highly extended condition typical of a complete interphase. Cytokinesis can partition the cell into two haploid products, forming a dyad. Statement IV belongs to anaphase II rather than telophase I. Sister chromatids remain joined after the first division, and each telophase-I chromosome is still replicated. The valid statements describe a transitional state in which chromosome-set reduction has occurred, cellular products may form and the chromosomes remain prepared for the second meiotic division. Telophase I can re-establish nuclei around replicated haploid chromosome sets; sister-chromatid separation remains reserved for meiosis II.
215. Use the cellular arrangement described here. Two chromosome groups lie at opposite poles. Each chromosome still consists of two sister chromatids. Nuclear envelopes are beginning to surround the groups, and a cytoplasmic partition is forming between them. The stage is:
ⓐ. metaphase I
ⓑ. telophase I
ⓒ. telophase II
ⓓ. anaphase II
Correct Answer: telophase I
Explanation: The chromosome groups have already separated to opposite poles, eliminating metaphase I as the stage. Each chromosome remains composed of two sister chromatids, showing that the centromeres did not divide during the preceding anaphase. Nuclear-envelope formation and beginning cytoplasmic separation identify telophase. Since replicated chromosomes are being enclosed after homologue segregation, the stage is telophase I rather than telophase II. In telophase II, the sister chromatids would already have separated and each chromosome would be represented by a single chromatid. The forming cytoplasmic partition may produce two haploid cells, but each cell will retain chromosomes at the \(n,2C\) condition until meiosis II separates the sisters. Each pole receives one member of every homologous pair, so ploidy falls to \(n\) even though each chromosome still contains two sister chromatids. The metaphase-I arrangement directs homologues towards opposite poles, while the joined sister chromatids are retained for separation in meiosis II.
216. A diploid meiocyte produces two cells. Each cell has one chromosome from every homologous pair, but every chromosome still contains two sister chromatids. Nuclear membranes are present, and the chromosomes remain partly condensed. These cells are:
ⓐ. telophase-I products forming a haploid dyad
ⓑ. mitotic daughter cells in the \(2n,2C\) state
ⓒ. telophase-II products forming a haploid tetrad
ⓓ. premeiotic cells entering the \(2n,4C\) state
Correct Answer: telophase-I products forming a haploid dyad
Explanation: Each cell has one member of every homologous pair, so the reduction from \(2n\) to \(n\) has already occurred. The chromosomes still possess two sister chromatids, indicating that meiosis II has not yet separated them. The presence of nuclear membranes and partly condensed chromosome sets is consistent with telophase I. Cytokinesis after this stage produces two haploid cells known collectively as a dyad. Mitotic daughters would retain the diploid chromosome complement, while telophase-II products would usually form four cells whose chromosomes contain one chromatid each. The combined evidence of product number, ploidy, replication state and nuclear organisation identifies the products of the first meiotic division rather than merely any pair of newly formed cells.
217. Four cellular intervals are compared below.
| Interval | Ploidy and DNA state | DNA synthesis | Position in division sequence |
|---|
| P | \(2n,2C\) | No | Before premeiotic S phase |
| Q | \(2n,4C\) | Completed | Immediately before meiosis I |
| R | \(n,2C\) | No | Between meiosis I and meiosis II |
| S | \(n,1C\) | No | After meiosis II |
Which ordered pair demonstrates the reductional change produced by meiosis I without an intervening round of DNA replication?
ⓐ. P → Q
ⓑ. R → S
ⓒ. Q → R
ⓓ. P → S
Correct Answer: Q → R
Explanation: Interval Q represents the replicated diploid meiocyte immediately before meiosis I, with the condition \(2n,4C\). During the first meiotic division, homologous chromosomes are allocated to different products while sister chromatids remain joined. Each product enters the between-division interval with the condition \(n,2C\), represented by R. The change Q → R records both the reduction in chromosome-set number and the partitioning of DNA from \(4C\) to \(2C\) per cell. The table also states that no DNA synthesis occurs in R, which identifies the interval as interkinesis rather than a second S phase. P → Q represents premeiotic replication, while R → S represents sister-chromatid separation in meiosis II. The ordered pair Q → R captures the specific reductional outcome of meiosis I and the unchanged replicated state carried into interkinesis. No second replication intervenes between those two conditions. The result records a halving of homologous chromosome sets while each retained chromosome still carries two sister chromatids.
218. Meiocytes are supplied with a labelled DNA precursor only during the interval between meiosis I and meiosis II. The label is not incorporated into nuclear chromosomes, yet the cells later complete meiosis II normally. The strongest inference is:
ⓐ. interkinesis normally lacks DNA replication
ⓑ. homologues remain paired throughout interkinesis
ⓒ. chromosome number returns from \(n\) to \(2n\)
ⓓ. sister chromatids separate before prophase II
Correct Answer: interkinesis normally lacks DNA replication
Explanation: Incorporation of a labelled DNA precursor would indicate synthesis of new DNA. The absence of nuclear labelling during the interval between the two meiotic divisions shows that no additional genome replication occurred. The later completion of meiosis II demonstrates that the second division can proceed using the sister chromatids formed during the single premeiotic S phase. Homologues have already separated in meiosis I and do not remain paired during interkinesis. Ploidy stays haploid rather than returning to \(2n\), and sister chromatids remain associated until anaphase II. The experimental evidence directly supports the defining feature of interkinesis: it is an interval between divisions, but unlike a complete interphase, it contains no S phase.
219. A graph plots DNA content per cell against meiotic progression. The curve falls from \(4C\) to \(2C\) after meiosis I, remains horizontal during interval P and falls to \(1C\) after meiosis II. Interval P represents:
ⓐ. interkinesis without DNA replication
ⓑ. a second S phase that restores \(4C\) DNA
ⓒ. pachytene with active crossing over
ⓓ. \(G_2\) before the first meiotic division
Correct Answer: interkinesis without DNA replication
Explanation: Meiosis I partitions the \(4C\) DNA of the replicated diploid meiocyte into two cells, each containing \(2C\). The horizontal segment during interval P shows that DNA content per cell remains unchanged. This is the expected pattern during interkinesis, where no replication occurs. Meiosis II then separates sister chromatids and reduces DNA content per final cell from \(2C\) to \(1C\). A second S phase would produce a rising segment rather than a plateau. Pachytene occurs before meiosis I while the cell still contains \(4C\) DNA, and \(G_2\) is also a predivision diploid interval. The graph demonstrates that the two meiotic divisions are separated by a pause without an intervening genome-duplication event. Axes, interval labels and curve shape must be interpreted together before assigning a stage. A change in slope marks a change in rate or process, whereas a horizontal segment represents persistence of the current state.
220. Assertion: Interkinesis includes a second S phase that prepares chromosomes for meiosis II.
Reason: DNA replication does not normally occur between meiosis I and meiosis II.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain the Assertion
ⓒ. Assertion is true and Reason is false, so Reason cannot correctly explain the Assertion
ⓓ. Assertion is false and Reason is true, so Reason cannot correctly explain the Assertion
Correct Answer: Assertion is false and Reason is true, so Reason cannot correctly explain the Assertion
Explanation: The assertion is false. Interkinesis is the interval between meiosis I and meiosis II, but it does not contain another S phase. The reason is true and directly states the defining replication rule. Each chromosome entering meiosis II already consists of two sister chromatids produced during the single premeiotic S phase. Meiosis I separates homologous chromosomes while retaining those sisters, leaving each interkinesis cell in the \(n,2C\) condition. A second replication would increase DNA content and disturb the normal pattern of one genome duplication followed by two divisions. Interkinesis may involve limited preparation for the next division, but that preparation must not be confused with renewed DNA synthesis. Interkinesis preserves the replicated chromosomes produced before meiosis I and carries them into the second division without another genome-duplication step.